So the question is in the title itself:
How do I simplify the absolute value of $\sqrt{1-\iota t}$ for any arbitrary $t\in\mathbb R$?
I would have replaced $\iota$ by $-\iota$, multiplied by it, and then taken the square root in usual circumstances, however, I am stuck on the case when $|t|>1$, because doing the same in this case would be a problem analogous to $-2 = \sqrt2\iota\times\sqrt2\iota= \sqrt{-2}\times\sqrt{-2}=\sqrt{4}=2$.
What to do in this case?
Note that $|1-it|=\sqrt{1+t^2}$. The number $1-it$ has two square roots, but if $r$ is one such square root, then$$|r|=\sqrt{|1-it|}=\sqrt[4]{1+t^2}.$$